In the triangle ABC, the three angles satisfy 2A = B + C, and the maximum and minimum edges are the two roots of the equation 3x ＾ 2-27 + 32 = 0, respectively Then a = It's 27x, not 27

In the triangle ABC, the three angles satisfy 2A = B + C, and the maximum and minimum edges are the two roots of the equation 3x ＾ 2-27 + 32 = 0, respectively Then a = It's 27x, not 27

Please complete the title and ask the two roots of 3x ＾ 2-27? + 32 = 0

In the triangle ABC, we know that (A's square + B's Square) * sin (a-b) = (A's Square - B's Square) * sin (a + b), and prove that the triangle ABC is isosceles triangle

(A's square + B's Square) * sin (a-b) = (A's Square - B's Square) * sin (a + b) (a ^ 2 + B ^ 2) / (a ^ 2-B ^ 2) = (sinacosb + cosasinb) / (sinacosb cosasinb) a ^ 2 / b ^ 2 = sinacosb / cosasinb from the sine theorem, (Sina) ^ 2 / (SINB) ^ 2 = sinacosb / cosasinbsina / SINB = CoSb /

It is known that a, B and C are the three sides of △ ABC. The values of a and B are the two real roots of the equation x ^ 2 - (4-C) x + 4C + 8 = 0 and satisfy 25sina = 9C,
What is the perimeter of △ ABC?

a+b=-(4-c)
ab=4c+8
So a ^ 2 + B ^ 2 = (a + b) ^ 2-2ab = (4-C) ^ 2-2 (4C + 8) = C ^ 2-16c
From the sine formula Sina / a = sinc / C, so Sina = a * sinc / C
Sina = a * sinc / C is replaced by 25sina = 9C to get sinc = 9C ^ 2 / 25A
The cosine formula COSC = (a ^ 2 + B ^ 2-C ^ 2) / 2Ab = - 16C / 2 (4C + 8) = - 2C / (c + 2)
Finally, sinc ^ 2 + COSC ^ 2 = 1 to get C
The rest is simple

It is known that the three sides of △ ABC are a, B, C, where a and B are two of the equations X2 - (c + 4) x + 4C + 8 = 0,
Prove that (1) △ ABC is a right triangle, (2) if a = B, find the length of the three sides of the triangle

a2+b2=(c+4)2-2(4c+8)
=c2
So it's a right triangle
A = b = 2 radical 2 C = 4

It is known that in △ ABC, the opposite sides of ∠ a, B and C are a, B and C respectively, if a and B are the two roots of the quadratic equation X2 - (c + 4) x + 4C + 8 = 0 with respect to x, and 9C = 25A · Sina. (1) prove that △ ABC is a right triangle. (2) find the length of three sides of △ ABC

(1) It is proved that: ∵ A and B are the two roots of the quadratic equation X2 - (c + 4) x + 4C + 8 = 0 with respect to X ∵ a + B = C + 4, ab = 4C + 8, ∵ A2 + B2 = (c + 4) 2-2 (4C + 8) = C2 ∵ ABC is a right triangle; (2) in RT △ ABC, Sina = AC, ∵ 9C = 25A ∵ Sina, ∵ 25a2 = 9c2, ∵ a = 35C is obtained by Pythagorean theorem: B = 45C, substituting a = 35C, B = 45C into a + B = C + 4, 75c = C + 4, and the solution is C = 10,